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1Formulas and methods
Three-phase power
Y connection
VL = √3·Vph, IL = Iph
Δ connection
VL = Vph, IL = √3·Iph
Total real power
P = √3·VL·IL·cos φ = 3·Iph²·R
Apparent and reactive power
S = √3·VL·IL, Q = P·tan φ
Correction capacitors
QC = P(tan φ1 − tan φ2), C = QC/(3ωV²)
Voltage across each capacitor
Δ: VL Y: VL/√3
Motor input
Pin = Pout/η
Transformers
Turns ratio
a = N1/N2 = V1/V2
Current ratio
I1/I2 = N2/N1 = 1/a
Rated current
I = S/V
Referred impedance
Z1′ = a²·ZL
Copper loss at part load
Pcu = x²·Pcu,fl
Efficiency
η = x·S·pf/(x·S·pf + Pcore + x²·Pcu,fl)
Maximum efficiency
x = √(Pcore/Pcu,fl)
AC power & power factor
Real power
P = V·I·cos φ
Reactive power
Q = V·I·sin φ
Apparent power
S = V·I = √(P² + Q²)
Power factor
pf = cos φ = P/S
Correction capacitor
QC = P(tan φ1 − tan φ2), C = QC/(ωV²)
AC impedance & phasors
Impedances
ZR = R, ZL = jωL, ZC = 1/(jωC)
Angular frequency
ω = 2πf
Magnitude and angle
|Z| = √(R² + X²), φ = atan(X/R)
Phasor Ohm’s law
V = I·Z
Method
Convert the frequency: ω = 2πf.
Find each reactance: XL = ωL and XC = 1/(ωC).
Add impedances like resistors, with Z = R + jX.
Find the magnitude |Z| = √(R² + X²) and the angle φ = atan(X/R), then use Ohm’s law with phasors.
DC motors
Armature circuit
V = E + IaRa
Back-EMF and torque
E = k·ω, T = k·Ia
Speed
ω = 2π·n/60, ω = (V − IaRa)/k
No-load speed and stall torque
ω0 = V/k, Tstall = k·V/Ra
Powers
Pin = V·Ia, Pdev = E·Ia = T·ω, Pout = Pdev − Prot
Starting
Istart = V/(Ra + Rst)
Speed regulation
SR = (n0 − nFL)/nFL
Maximum developed power
Pmax = V²/(4Ra)
Chopper drive
Va = D·Vdc
Method
Draw the armature circuit: supply V, resistance Ra and the back-EMF E in series.
Use V = E + Ia·Ra to link voltage, current and back-EMF.
Use E = k·ω for speed and T = k·Ia for torque, with ω = 2π·n/60.
For power and efficiency, start from V·Ia, subtract the copper loss Ia²·Ra and then the rotational losses.
2Practice problems
1.Three-phase powerMedium
A balanced Y-connected load with an impedance of 15 + j20 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 480 V. Find the total real power P absorbed by the load in kW.
VL = 480 VZ = 15 + j20 Ω per phase
P = kW
2.Three-phase powerHard
A balanced three-phase load takes 150 kW at a power factor of 0.8 lagging from a 690 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.92 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.
P = 150 kWpf1 = 0.8pf2 = 0.92VL = 690 Vf = 50 Hz
QC = kvar
3.TransformersMedium
An amplifier with an output resistance of 1152 Ω drives an 8 Ω loudspeaker through an ideal transformer. For maximum power transfer the loudspeaker must look like 1152 Ω from the amplifier. Find the turns ratio a = N₁/N₂.
Rs = 1152 ΩRL = 8 Ω
a =
4.TransformersHard
A single-phase 25 kVA transformer has a core loss of 75 W and a copper loss of 187.5 W at rated load. It supplies 50% of its rated kVA at a power factor of 0.8 lagging. Find the total losses Ploss in watts at this load.
A single-phase motor takes 28 kW at a power factor of 0.7 lagging from a 240 V supply. Find the line current I.
P = 28 kWpf = 0.7V = 240 V
I = A
6.AC power & power factorHard
A 200 kVA single-phase transformer supplies a load of 93 kW at a power factor of 0.75 lagging. How much extra real power ΔP can be added as a resistive (unity power factor) load before the transformer reaches its kVA rating?
Smax = 200 kVAP1 = 93 kWpf1 = 0.75
ΔP = kW
7.AC impedance & phasorsMedium
A series RLC circuit is connected to a 24 V (rms), 500 Hz supply. Find the real power P taken from the supply.
V = 24 V (rms)f = 500 HzR = 100 ΩL = 100 mHC = 2.2 µF
P = W
8.AC impedance & phasorsHard
A series RLC circuit is connected to a 10 V (rms), 60 Hz supply. Find the resonant frequency f0 of the circuit.
V = 10 V (rms)f = 60 HzR = 10 ΩL = 250 mHC = 47 µF
f0 = Hz
9.DC motorsMedium
A DC motor with k = 0.8 V·s/rad runs from 220 V. Neglecting all losses at no load, find its no-load speed n0 in rpm.
V = 220 Vk = 0.8 V·s/rad
n0 = rpm
10.DC motorsHard
A separately excited DC motor (k = 2 V·s/rad at full field, Ra = 1.2 Ω, V = 240 V) drives a constant load torque of 20 N·m. The field flux is reduced to 75 % of its full value. Find the new speed n.
Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ.
Phase voltage
Vph = 480/√3 = 277.1 V
Phase current
|Z| = √(15² + 20²) = 25 Ω
Iph = 277.1/25 = 11.09 A
Total real power
Only the resistive part of each impedance absorbs real power.
P = 3·Iph²·R = 3·11.09²·15 = 5.53 kW
Answer
P = 5.53 kW
Check: cos φ = R/|Z| = 0.6, and √3·VL·IL·cos φ gives the same power.
2.Three-phase power
Answer: QC = 48.6 kvar
Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ. Power factor correction: capacitors supply reactive power locally, QC = P·(tan φ1 − tan φ2). Each of the three capacitors supplies QC/3 = V²·ω·C, where V is the voltage across that capacitor: VL in a Δ bank, VL/√3 in a Y bank.
Reactive power before and after
The real power is unchanged, so each power factor fixes the reactive power: Q = P·tan φ.
Q1 = 150·tan(acos 0.8) = 150·0.75 = 112.5 kvar
Q2 = 150·tan(acos 0.92) = 150·0.426 = 63.9 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 48.6 kvar
Answer
QC = 48.6 kvar
3.Transformers
Answer: a = 12
Ideal transformer: the voltages are in the ratio of the turns and the currents in the inverse ratio: V₁/V₂ = N₁/N₂ = a and I₁/I₂ = N₂/N₁ = 1/a, so the power in equals the power out, V₁·I₁ = V₂·I₂. A load impedance Z on the secondary looks like a²·Z from the primary.
Match the referred load
a²·8 = 1152
a = √(1152/8) = √144 = 12
Answer
a = 12
4.Transformers
Answer: Ploss = 121.9 W
Transformer losses: the core loss (hysteresis and eddy currents) depends on the voltage and is the same at any load. The copper loss in the windings goes with the current squared, so at a fraction x of rated load it is x²·Pcu,fl. Efficiency η = Pout/(Pout + Pcore + Pcu), with Pout = x·S·pf. The efficiency is highest at the load where the copper loss equals the core loss.
Copper loss at this load
Pcu = x²·Pcu,fl = 0.25·187.5 = 46.88 W
Total losses
The core loss is the same at any load.
Ploss = 75 + 46.88 = 121.9 W
Answer
Ploss = 121.9 W
The efficiency at this load is η = 98.8%.
5.AC power & power factor
Answer: I = 166.7 A
AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.
Line current
Only the real part of the power does work, so divide by the power factor.
I = P/(V·cos φ) = 28000/(240·0.7) = 166.7 A
Answer
I = 166.7 A
6.AC power & power factor
Answer: ΔP = 89.41 kW
AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.
Reactive power of the existing load
Q1 = 93·tan(arccos 0.75) = 82.02 kvar
Largest total real power
At full rating S = Smax, and Q is unchanged by the resistive load.
P = √(Smax² − Q1²) = √(200² − 82.02²) = 182.4 kW
Extra real power
ΔP = 182.4 − 93 = 89.41 kW
Answer
ΔP = 89.41 kW
7.AC impedance & phasors
Answer: P = 1.488 W
AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.
Reactances
XL = 314.2 Ω, XC = 144.7 Ω
Impedance
|Z| = √(100² + (169.5)²) = 196.8 Ω
Real power
Inductors and capacitors only store and return energy; the resistor absorbs all of the real power.
I = 24/196.8 = 0.122 A
P = I²·R = 0.122²·100 = 1.488 W
Answer
P = 1.488 W
8.AC impedance & phasors
Answer: f0 = 46.43 Hz
AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.
Resonance
At resonance the inductive and capacitive reactances cancel: ωL = 1/(ωC). The supply frequency does not matter.
DC motor: the armature is a resistance Ra in series with the back-EMF E, so V = E + Ia·Ra. The back-EMF grows with speed, E = k·ω, and the torque with current, T = k·Ia, with the same motor constant k (V·s/rad = N·m/A) in SI units. Convert speeds with ω = 2π·n/60. The developed power is E·Ia = T·ω; the input power is V·Ia, and the copper loss Ia²Ra and the rotational loss make the difference. The speed falls in a straight line with torque, ω = V/k − Ra·T/k², from the no-load speed V/k to zero at the stall torque k·V/Ra. At standstill E = 0, so the starting current V/Ra is large unless a resistor limits it.
No-load speed
ω₀ = V/k = 220/0.8 = 275 rad/s
n0 = 60·275/(2π) = 2626 rpm
Answer
n0 = 2626 rpm
10.DC motors
Answer: n = 1426 rpm
DC motor: the armature is a resistance Ra in series with the back-EMF E, so V = E + Ia·Ra. The back-EMF grows with speed, E = k·ω, and the torque with current, T = k·Ia, with the same motor constant k (V·s/rad = N·m/A) in SI units. Convert speeds with ω = 2π·n/60. The developed power is E·Ia = T·ω; the input power is V·Ia, and the copper loss Ia²Ra and the rotational loss make the difference. The speed falls in a straight line with torque, ω = V/k − Ra·T/k², from the no-load speed V/k to zero at the stall torque k·V/Ra. At standstill E = 0, so the starting current V/Ra is large unless a resistor limits it.